Power Series Expansion for Real Area Hyperbolic Secant
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Theorem
The (real) area hyperbolic secant function has a Taylor series expansion:
\(\ds \arsech x\) | \(=\) | \(\ds \ln \frac 2 x - \paren {\sum_{n \mathop = 1}^\infty \frac {\paren {2 n}! x^{2 n} } {2^{2 n} \paren {n!}^2 \paren {2 n} } }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \ln \frac 2 x - \paren {\dfrac 1 2 \dfrac {x^2} 2 + \dfrac {1 \times 3} {2 \times 4} \dfrac {x^4} 4 + \dfrac {1 \times 3 \times 5} {2 \times 4 \times 6} \dfrac {x^6} 6 + \cdots}\) |
for $0 < x \le 1$.
Proof
From Power Series Expansion for Real Area Hyperbolic Cosine:
\(\ds \arcosh x\) | \(=\) | \(\ds \ln 2 x - \paren {\sum_{n \mathop = 1}^\infty \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2 \paren {2 n} x^{2 n} } }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \ln 2 x - \paren {\dfrac 1 2 \dfrac 1 {2 x^2} + \dfrac {1 \times 3} {2 \times 4} \dfrac 1 {4 x^4} + \dfrac {1 \times 3 \times 5} {2 \times 4 \times 6} \dfrac 1 {6 x^6} + \cdots}\) |
for $x \ge 1$.
From Real Area Hyperbolic Cosine of Reciprocal equals Real Area Hyperbolic Secant:
- $\map \arcosh {\dfrac 1 x} = \arsech x$
So:
\(\ds \arsech x\) | \(=\) | \(\ds \ln \frac 2 x - \paren {\sum_{n \mathop = 1}^\infty \frac {\paren {2 n}! x^{2 n} } {2^{2 n} \paren {n!}^2 \paren {2 n} } }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \ln \frac 2 x - \paren {\dfrac 1 2 \dfrac {x^2} 2 + \dfrac {1 \times 3} {2 \times 4} \dfrac {x^4} 4 + \dfrac {1 \times 3 \times 5} {2 \times 4 \times 6} \dfrac {x^6} 6 + \cdots}\) |
For $1 \le \dfrac 1 x$ we have that $0 < x \le 1$.
Hence the result.
$\blacksquare$